(a) Differentiate \(r^2 = e^{\sin \theta} \cos \theta\) with respect to \(\theta\) and set \(\frac{dr}{d\theta} = 0\). This gives:

\(-e^{\sin \theta} \sin \theta + e^{\sin \theta} \cos^2 \theta = 0\)

\(\cos^2 \theta = \sin \theta\)

\(1 - \sin^2 \theta = \sin \theta\)

\(\sin \theta = \frac{1}{2}(-1 + \sqrt{5})\)

\(\theta = 0.666\)

\(r = \sqrt{e^{\sin(0.666)} \cos(0.666)} = 1.208\)

(b) Use \(x = e^{\sin \theta} \cos \theta\) and differentiate with respect to \(\theta\):

\(\frac{dx}{d\theta} = -\frac{1}{2}e^{\sin \theta} \cos \theta \sin \theta + e^{\sin \theta} \cos \theta \cos \theta - \frac{1}{2}e^{\sin \theta} \sin \theta \cos \theta = 0\)

\(-3\sin \theta + \cos^2 \theta = 0\)

\(-3\sin \theta + 1 - \sin^2 \theta = 0\)

\(\sin \theta = \frac{1}{2}(-3 + \sqrt{13})\)

\(\theta = 0.308\)

\(r = \sqrt{e^{\sin(0.308)} \cos(0.308)} = 1.136\)

(c) The sketch shows a closed loop passing through O and one other point on the initial line. It does not go into the 2nd and 3rd quadrants and is more in the 1st than the 4th quadrant.

(d) The area is given by:

\(\frac{1}{2} \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi} e^{\sin \theta} \cos \theta \, d\theta\)

\(= \frac{1}{2} \left[ e^{\sin \theta} \right]_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\)

\(= \frac{1}{3}(e - e^{-1})\)

(a) We start with \(\sum_{r=1}^{n} (2r+1) = 2\sum_{r=1}^{n} r + n\). Using the formula for the sum of the first \(n\) natural numbers, \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\), we have:

\(2\sum_{r=1}^{n} r + n = 2 \times \frac{n(n+1)}{2} + n = n(n+1) + n = n(n+2)\).

(b) To show \(\frac{2r+1}{(r^2+1)(r^2+2r+2)} = \frac{1}{r^2+1} - \frac{1}{r^2+2r+2}\), we find a common denominator:

\(\frac{1}{r^2+1} - \frac{1}{r^2+2r+2} = \frac{r^2+2r+2 - (r^2+1)}{(r^2+1)(r^2+2r+2)} = \frac{2r+1}{(r^2+1)(r^2+2r+2)}\).

(c) Using the result from (b), the series becomes a telescoping series:

\(\sum_{r=1}^{n} \left( \frac{1}{r^2+1} - \frac{1}{r^2+2r+2} \right) = \frac{1}{2} - \frac{1}{n^2+2n+2}\).

(d) As \(n \to \infty\), the term \(\frac{1}{n^2+2n+2} \to 0\), so the sum converges to \(\frac{1}{2}\).

First, find the first derivative:

\(\frac{d}{dx}(x \ln x) = 1 + \ln x\).

Check the base case for \(n = 2\):

\(\frac{d^2}{dx^2}(x \ln x) = x^{-1} = (-1)^2 (2-2)! x^{1-2}\).

Assume that \(\frac{d^k}{dx^k}(x \ln x) = (-1)^k (k-2)! x^{1-k}\) holds for some integer \(k \geq 2\).

Differentiate the \(k\)-th derivative:

\(\frac{d^{k+1}}{dx^{k+1}}(x \ln x) = (-1)^k (k-2)!(1-k)x^{1-k-1}\).

Simplify:

\(= (-1)^{k+1} ((k+1)-2)! x^{1-(k+1)}\).

Thus, \(\frac{d^n}{dx^n}(x \ln x) = (-1)^n (n-2)! x^{1-n}\) is also true for \(n = k+1\).

By induction, the statement is true for every integer \(n \geq 2\).

(a) To find the equation of the plane, we first determine the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (-5\mathbf{i} + 3\mathbf{j} + \mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = -5\mathbf{i} + \mathbf{j} - 2\mathbf{k}\)

\(\overrightarrow{AC} = (\mathbf{i} + 2\mathbf{j} + 5\mathbf{k}) - (2\mathbf{j} + 3\mathbf{k}) = \mathbf{i} + 2\mathbf{k}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -5 & 1 & -2 \\ 1 & 0 & 2 \end{vmatrix} = 2\mathbf{i} + 8\mathbf{j} - \mathbf{k}\)

Using point \(A(0, 2, 3)\), the equation of the plane is:

\(2(0) + 8(2) - 1(3) = 13 \Rightarrow 2x + 8y - z = 13\)

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|2(0) + 8(0) - 1(0) - 13|}{\sqrt{2^2 + 8^2 + (-1)^2}} = \frac{13}{\sqrt{69}} = 1.57\)

(c) The acute angle \(\theta\) between the line \(OA\) and the plane is found using the dot product:

\(\cos \theta = \frac{\mathbf{n} \cdot \overrightarrow{OA}}{\|\mathbf{n}\| \|\overrightarrow{OA}\|}\)

\(\overrightarrow{OA} = 2\mathbf{j} + 3\mathbf{k}\)

\(\mathbf{n} \cdot \overrightarrow{OA} = 2(0) + 8(2) - 1(3) = 13\)

\(\|\mathbf{n}\| = \sqrt{69}, \quad \|\overrightarrow{OA}\| = \sqrt{13}\)

\(\cos \theta = \frac{13}{\sqrt{69} \sqrt{13}}\)

\(\theta = \cos^{-1}\left(\frac{13}{\sqrt{69} \sqrt{13}}\right) = 0.449\) radians

(a) The determinant of the matrix \(\begin{pmatrix} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{pmatrix}\) is calculated as:

\(\left| \begin{array}{ccc} 1 & \alpha & \beta \\ \alpha & 1 & \gamma \\ \beta & \gamma & 1 \end{array} \right| = 1 \cdot \left| \begin{array}{cc} 1 & \gamma \\ \gamma & 1 \end{array} \right| - \alpha \cdot \left| \begin{array}{cc} \alpha & \gamma \\ \beta & 1 \end{array} \right| + \beta \cdot \left| \begin{array}{cc} \alpha & 1 \\ \beta & \gamma \end{array} \right|\)

\(= 1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma\)

Since the matrix is singular, the determinant is zero:

\(1 - \alpha^2 - \beta^2 - \gamma^2 + 2\alpha\beta\gamma = 0\)

Given \(\alpha\beta\gamma = 1\), we have:

\(\alpha^2 + \beta^2 + \gamma^2 = 1 + 2(1) = 3\)

(b) We use the identity:

\(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)

Given \(\alpha^2 + \beta^2 + \gamma^2 = 3\), we have:

\(3 = b^2 - 2c\)

Also, given \(\alpha^3 + \beta^3 + \gamma^3 = 3\), we use:

\(\alpha^3 + \beta^3 + \gamma^3 = -b(3) - c(-b) + 3\)

\(3 = -3b + cb + 3\)

Solving these simultaneous equations, we find:

\(c = 3, \ b = 3\)